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Solving Quadratic Equations and Graphing Parabolas”, Chapter 9 from the Book Beginning Algebra (Index.Html) (V This is “Solving Quadratic Equations and Graphing Parabolas”, chapter 9 from the book Beginning Algebra (index.html) (v. 1.0). This book is licensed under a Creative Commons by-nc-sa 3.0 (http://creativecommons.org/licenses/by-nc-sa/ 3.0/) license. See the license for more details, but that basically means you can share this book as long as you credit the author (but see below), don't make money from it, and do make it available to everyone else under the same terms. This content was accessible as of December 29, 2012, and it was downloaded then by Andy Schmitz (http://lardbucket.org) in an effort to preserve the availability of this book. Normally, the author and publisher would be credited here. However, the publisher has asked for the customary Creative Commons attribution to the original publisher, authors, title, and book URI to be removed. Additionally, per the publisher's request, their name has been removed in some passages. More information is available on this project's attribution page (http://2012books.lardbucket.org/attribution.html?utm_source=header). For more information on the source of this book, or why it is available for free, please see the project's home page (http://2012books.lardbucket.org/). You can browse or download additional books there. i Chapter 9 Solving Quadratic Equations and Graphing Parabolas 1425 Chapter 9 Solving Quadratic Equations and Graphing Parabolas 9.1 Extracting Square Roots LEARNING OBJECTIVE 1. Solve quadratic equations by extracting square roots. Extracting Square Roots Recall that a quadratic equation is in standard form1 if it is equal to 0: where a, b, and c are real numbers and a ≠ 0. A solution to such an equation is called a root2. Quadratic equations can have two real solutions, one real solution, or no real solution. If the quadratic expression on the left factors, then we can solve it by factoring. A review of the steps used to solve by factoring follow: Step 1: Express the quadratic equation in standard form. Step 2: Factor the quadratic expression. Step 3: Apply the zero-product property and set each variable factor equal to 0. Step 4: Solve the resulting linear equations. For example, we can solve x 2 − 4 = 0 by factoring as follows: 1. Any quadratic equation in the form ax 2 + bx + c = 0, where a, b, and c are real numbers and a ≠ 0. 2. A solution to a quadratic equation in standard form. 1426 Chapter 9 Solving Quadratic Equations and Graphing Parabolas The two solutions are −2 and 2. The goal in this section is to develop an alternative method that can be used to easily solve equations where b = 0, giving the form The equation x 2 − 4 = 0 is in this form and can be solved by first isolating x 2 . If we take the square root of both sides of this equation, we obtain the following: Here we see that x = −2 and x = 2 are solutions to the resulting equation. In general, this describes the square root property3; for any real number k, The notation “±” is read “plus or minus” and is used as compact notation that ⎯⎯ ⎯⎯ indicates two solutions. Hence the statement x = ±√k indicates that x = −√k or ⎯⎯ x = √k. Applying the square root property as a means of solving a quadratic equation is called extracting the roots4. 3. For any real number k, if ⎯⎯ Example 1: Solve: x 2 . x 2 = k, then x = ±√k . − 25 = 0 4. Applying the square root property as a means of solving Solution: Begin by isolating the square. a quadratic equation. 9.1 Extracting Square Roots 1427 Chapter 9 Solving Quadratic Equations and Graphing Parabolas Next, apply the square root property. Answer: The solutions are −5 and 5. The check is left to the reader. Certainly, the previous example could have been solved just as easily by factoring. However, it demonstrates a technique that can be used to solve equations in this form that do not factor. Example 2: Solve: x 2 − 5 = 0. Solution: Notice that the quadratic expression on the left does not factor. We can extract the roots if we first isolate the leading term, x 2 . Apply the square root property. 9.1 Extracting Square Roots 1428 Chapter 9 Solving Quadratic Equations and Graphing Parabolas For completeness, check that these two real solutions solve the original quadratic equation. Generally, the check is optional. ⎯⎯ ⎯⎯ Answer: The solutions are −√5 and √5. Example 3: Solve: 4x 2 − 45 = 0. Solution: Begin by isolating x 2 . Apply the square root property and then simplify. 9.1 Extracting Square Roots 1429 Chapter 9 Solving Quadratic Equations and Graphing Parabolas Answer: The solutions are 3√5 and 3√5. − 2 2 Sometimes quadratic equations have no real solution. Example 4: Solve: x 2 + 9 = 0. Solution: Begin by isolating x 2 . After applying the square root property, we are left with the square root of a negative number. Therefore, there is no real solution to this equation. Answer: No real solution Reverse this process to find equations with given solutions of the form ±k. 9.1 Extracting Square Roots 1430 Chapter 9 Solving Quadratic Equations and Graphing Parabolas ⎯⎯ ⎯⎯ Example 5: Find an equation with solutions −2√3 and 2√3. Solution: Begin by squaring both sides of the following equation: Lastly, subtract 12 from both sides and present the equation in standard form. Answer: x 2 − 12 = 0 Try this! Solve: 9x 2 − 8 = 0. Answer: x 2√2 or x 2√2 = − 3 = 3 Video Solution (click to see video) Consider solving the following equation: To solve this equation by factoring, first square x + 2 and then put it in standard form, equal to zero, by subtracting 25 from both sides. 9.1 Extracting Square Roots 1431 Chapter 9 Solving Quadratic Equations and Graphing Parabolas Factor and then apply the zero-product property. The two solutions are −7 and 3. When an equation is in this form, we can obtain the solutions in fewer steps by extracting the roots. Example 6: Solve: (x + 2)2 = 25. Solution: Solve by extracting the roots. At this point, separate the “plus or minus” into two equations and simplify each individually. 9.1 Extracting Square Roots 1432 Chapter 9 Solving Quadratic Equations and Graphing Parabolas Answer: The solutions are −7 and 3. In addition to fewer steps, this method allows us to solve equations that do not factor. Example 7: Solve: (3x + 3)2 − 27 = 0. Solution: Begin by isolating the square. Next, extract the roots and simplify. Solve for x. 9.1 Extracting Square Roots 1433 Chapter 9 Solving Quadratic Equations and Graphing Parabolas ⎯⎯ ⎯⎯ Answer: The solutions are −1 − √3 and −1 + √3. Example 8: Solve: 9(2x − 1)2 − 8 = 0. Solution: Begin by isolating the square factor. Apply the square root property and solve. 9.1 Extracting Square Roots 1434 Chapter 9 Solving Quadratic Equations and Graphing Parabolas 3−2 2 3+2 2 Answer: The solutions are √ and √ . 6 6 2 Try this! Solve: 3(x − 5) − 2 = 0. 15± 6 Answer: √ 3 Video Solution (click to see video) Example 9: The length of a rectangle is twice its width. If the diagonal measures 2 feet, then find the dimensions of the rectangle. 9.1 Extracting Square Roots 1435 Chapter 9 Solving Quadratic Equations and Graphing Parabolas Solution: The diagonal of any rectangle forms two right triangles. Thus the Pythagorean theorem applies. The sum of the squares of the legs of a right triangle is equal to the square of the hypotenuse: Solve. 9.1 Extracting Square Roots 1436 Chapter 9 Solving Quadratic Equations and Graphing Parabolas Here we obtain two solutions, w 2 and w 2 . Since the problem asked for a = − √5 = √5 length of a rectangle, we disregard the negative answer. Furthermore, we will rationalize the denominator and present our solutions without any radicals in the denominator. Back substitute to find the length. Answer: The length of the rectangle is 4√5 feet and the width is 2√5 feet. 5 5 KEY TAKEAWAYS • Solve equations of the form ax 2 + c = 0by extracting the roots. • Extracting roots involves isolating the square and then applying the square root property. After applying the square root property, you have two linear equations that each can be solved. Be sure to simplify all radical expressions and rationalize the denominator if necessary. 9.1 Extracting Square Roots 1437 Chapter 9 Solving Quadratic Equations and Graphing Parabolas TOPIC EXERCISES Part A: Extracting Square Roots Solve by factoring and then solve by extracting roots. Check answers. 1. x 2 − 36 = 0 2. x 2 − 81 = 0 3. 4y 2 − 9 = 0 4. 9y 2 − 25 = 0 5. (x − 2)2 − 1 = 0 6. (x + 1)2 − 4 = 0 2 7. 4(y − 2) − 9 = 0 2 8. 9(y + 1) − 4 = 0 9. −3(t − 1)2 + 12 = 0 10. −2(t + 1)2 + 8 = 0 2 11. (x − 5) − 25 = 0 12. (x + 2)2 − 4 = 0 Solve by extracting the roots. 13. x 2 = 16 14. x 2 = 1 9.1 Extracting Square Roots 1438 Chapter 9 Solving Quadratic Equations and Graphing Parabolas 15. y 2 = 9 16. y 2 = 64 17. x 2 1 = 4 18. x 2 1 = 9 19. y 2 = 0.25 20. y 2 = 0.04 21. x 2 = 12 22. x 2 = 18 23. 16x 2 = 9 24. 4x 2 = 25 25. 2t2 = 1 26. 3t2 = 2 27. x 2 − 100 = 0 28. x 2 − 121 = 0 29. y 2 + 4 = 0 30. y 2 + 1 = 0 31.
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